The diagram below shows a ray of light passing from medium X into air.

What is the index of refraction of medium X?

Question

The diagram below shows a ray of light passing from medium X into air.

What is the index of refraction of medium X?

BlackTom AI · Accepted Answer

The question presents a refraction scenario: a ray travels from medium X into air and we’re given the Snell’s law relation n1 sin(theta1) = n2 sin(theta2). Here, theta1 (in medium X) is 30°, and theta2 (in air) is 60°. Since air’s index n2 is approximately 1.0, we can solve for n1.

Option-by-option analysis:

Option A: 1.73
- This value comes from applying Snell’s law: n1 sin(30°) = n2 sin(60°) with n2 ≈ 1.0. Sin(30°) = 0.5 and sin(60°) ≈ 0.866, so n1 ≈ 0.866 / 0.5 = 1.732, which rounds to 1.73. This matches the computed result and is therefore correct.

Option B: 0.577
- This is roughly 1 / 1.732. It would correspond to a much smaller refractive index, which would imply a much larger angle in the medium than observed, conflicting with the given 30° incidence and 60° refraction. It does not satisfy Snell’s law for the given angles.

Option C: 2.00
- If n1 were 2.00, Snell’s law would give sin(theta2) = n1/n2 sin(theta1) = 2.00 * sin(30°) = 2.00 * 0.5 = 1.0, which would imply theta2 = 90°, not the given 60°. This contradicts the diagram, so this option is incorrect.

Option D: 0.500
- An index of 0.500 is implausibly small for a real medium and would produce a refraction angle that does not align with the stated 60° in air when using Snell’s law with theta1 = 30°. It fails the given geometry and the fundamental relation, so it’s not feasible.

In summary, only 1.73 satisfies Snell’s law with theta1 = 30° in medium X and theta2 = 60° in air, using n2 ≈ 1.0 for air.

The diagram below shows a ray of light passing from medium X into air. What is the index of refraction of medium X?

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